Computing endpoints of a rectangle?

[Helper]

HI,
I have a picture control on the front panel of my VI. I need to be able to draw rectangles of different lengths and at different places during runtime.I'm almost certain that this is too complicated of a question to get an ideal answer to, but basically I was wondering if anyone had any ideas of some algorithm that I could use to figure out the endpoints of this "rectangle" if what I'm given during runtime is an angle where it should be relative to the center of the picture and the length that the rectangle should be? (The width is going to be constant). Any ideas will be appreciated,
Thanks

[CraigGraham]

You need to use some tedious trigonometry and break the rectangle into some
appropriate triangles to calculate the x and y offsets of the corners with
respect to the coordinates of the centre and the angle of rotation. I
started to do it on paper to scan in but made a mistake and I don't really
have time to go into such depth. Unless you want to contract me to write a
routine of course 🙂

What you're doing is a fairly simple task in vector graphics, which have
been around a long time; if you look up a web article, or even a computer
graphics textbook, they should tell you fairly early on how to do simple 2D
rotations.

--
Craig Graham
Physicist/Labview Programmer
Lancaster University, UK


"Helper" wrote in message
news:50650000000800000077390000-1
011517314000@exchange.ni.com...
> HI,
> I have a picture control on the front panel of my VI. I need to be
> able to draw rectangles of different lengths and at different places
> during runtime.I'm almost certain that this is too complicated of a
> question to get an ideal answer to, but basically I was wondering if
> anyone had any ideas of some algorithm that I could use to figure out
> the endpoints of this "rectangle" if what I'm given during runtime is
> an angle where it should be relative to the center of the picture and
> the length that the rectangle should be? (The width is going to be
> constant). Any ideas will be appreciated,
> Thanks

[Helper]

Thank you, I think that's the best response that I was hoping for.